This problem is most easily understood by reading my attached Gist: https://gist.github.com/KazW/e77e5c7603d7700d86c1
I have a simple encrypt/decrypt function in Ruby that works. Every time the encryption function is run, a new initialization vector is created using OpenSSL::Random.random_bytes(12). When I try to do the same steps in Elixir, using :crypto.strong_rand_bytes(12), it generates an IV that can be used for encryption, but causes decryption to fail.
Stranger still, if I use an IV generated in Ruby, and use it to Encrypt in Elixir. When passing the ciphertext to decrypt in Elixir, the decryption function returns the plaintext without error. I've researched the algorithm in use, the IV is supposed to be random bytes, the important factor is the number of bytes in it (12).
My initial thought is that perhaps I'm calling the wrong OpenSSL method to generate the IV. However, I wasn't able to find any information as to what the correct method would be.
The issue here is that you are using String functions to read the data, which only works with Unicode, and UTF8 characters are variable-width. You should use pattern matching instead, which uses single-byte width.
https://gist.github.com/asonge/4f035a38a9b31339d8f5
Related
Here's the problem, a string has been passed through three separate encryptions in the following order: Original -> Base64 -> AES-256 -> Blowfish (Keyless) -> Final. Write a method that takes this triple encoded string mystery_string = "OXbVgH7UriGqmRZcqOXUOvJt8Q4JKn5MwD1XP8bg9yHwhssYAKfWE+AMpr25HruA" and fully unencrypts it to its original state.
I looked into different libraries/documentation for aes256 and blowfish but all of them required a key. The only one that did not require a key was Base64 (i.e. Base64.encode64('some string') ). Not really sure where to go from here.
Firstly, the only way to crack AES-256 and Blowfish without the key is by brute force enumeration of every possibly 32-byte combination that could be used as the key. In theory, this means it's not crackable in our lifetime. There may be some vulnerabilities you could exploit as you also have the plain text, but I doubt you would have that in a real-life situation.
Second, and most importantly, just going by that site, encode-decode.comhttps://encode-decode.com/, you don't actually have enough information to decode the string even if you did know the password.
The various modes of operation for the AES256 cipher function requires either a 32-byte (or sometimes a 64-byte) key. The secret that you used (you may have just left it blank) needs to be converted into a 32-byte encryption key. This is done using a hashing algorithm, but we don't know which one is used. Hopefully, the site used a key derivation function, which provides several security benefits. However, key derivation functions require multiple parameters, and we would need to know what parameters to enter along with our secret to get the right encryption key.
Finally, we don't know if the secret is being concatenated with a salt before being hashed. Without knowing if a salt is used and what the salt is, we cannot determine the correct 32-byte key used to encrypt the plain text.
In summary, the answer to your question is: No, there is not a quick way to decrypt that string without knowing the key.
However, encryption is an awesome topic to learn.
I'd encourage you to look over the ruby docs for the OpenSSL library. They're actually quite good (besides the don'ts I mention below).
The PBKDF2 Password-based Encryption function is one of the key derivation functions I was referring to.
When encrypting with AES, you will most likely want to use AES-256-GCM which is authenticated encryption.
A couple of don'ts:
Don't use ciphers at random... understand their strengths and weaknesses
Don't use AES-128-EBC - explination
Another good encryption library is rb-NaCl.
In SQLCipher it is possible to provide a database key in two ways: either as a passphrase (with an associated number of PBKDF2 rounds) or as a raw key.
I am trying to find a way get the raw key (given as a 64 character hex string) given the passphrase (an arbitrary string) and number of rounds. I assume the solution lies in somehow performing the PBKDF2 key derivation myself, but I'm not too familiar with how this might work, or if it would produce a value that I could just feed as a raw key to PRAGMA KEY. In addition, the PBKDF2 function appears to take a salt, which I do not know. If I have to, I am sure I can dig into the sqlcipher source code to figure this out, but I'm hoping that there might already be a simple way to do this that I'm missing.
The details on SQLCipher key derivation can be found on the design page. The salt is stored in the first 16 bytes of the database file. You would then use the salt along with the passphrase with an appropriate number of PBKDF2-HMAC-SHA1 rounds (64,000 for SQLCipher 3, 4,000 for earlier versions)
How to decrypt a string with unknown encryption algorithm?
There is a string:
5aaC5p6c5L2g5a+55oiR5Lus5Zyo5YGa55qE5LqL5oOF5pyJ5YW06Laj77yM5bm25LiU5a+5cmFpbHMv5YmN56uv5byA5Y+R5pyJ6Ieq5L+h77yM5qyi6L+O5Y+R6YCB6YKu5Lu25YiwZ2hvc3RtNTVAZ2l0Y2FmZS5jb23pooTnuqbkuqTmtYHml7bpl7TvvIznoa7lrprkuYvlkI7lj6/ku6Xnm7TmjqXmnaXliLDmiJHku6znmoTlt6XkvZzlrqTlj4Lop4LkuqTmtYHvvIzosKLosKIK
I don't know the encryption algorithm. How to decrypt it?
To analyze and solve this problem, what should I learn?
It's not an encryption algorithm, it's base64. You can tell because of the +s.
http://www.opinionatedgeek.com/dotnet/tools/base64decode/
Try running it through this page, it'll turn into this:
如果你对我们在做的事情有兴趣，并且对rails/前端开发有自信，欢迎发送邮件到ghostm55#gitcafe.com预约交流时间，确定之后可以直接来到我们的工作室参观交流，谢谢
NOTE: If it was actually encrypted and you actually had no clue what it was encrypted with, you would be screwed, because any good encryption algorithm turns the output into meaningless gibberish, unusable without the key. Base64 has no key, you can just reverse it the same way every time.
This string appears to be a Base64 encoded string.
The decoded value is: 如果你对我们在做的事情有兴趣，并且对rails/前端开发有自信，欢迎发送邮件到ghostm55#gitcafe.com预约交流时间，确定之后可以直接来到我们的工作室参观交流，谢谢
Well, the string is likely Base64 encoded. If you decode it, you should get an effectively random piece of binary data if its encrypted (EDIT: As others have shown, it isn't encrypted, but the following would still apply if it were)
By checking the length, you can determine the block-size of the cipher. If its not an even block size, it likely could be a stream cipher (or a block cipher operated in stream mode).
However, any more information will need to be gleamed from other sources - as the point of good encryption is to make the data truly opaque.
Its Base 64 encryption.The above code is translated as:
如果你对我们在做的事情有兴趣，并且对rails/前端开发有自信，欢迎发送邮件到ghostm55#gitcafe.com预约交流时间，确定之后可以直接来到我们的工作室参观交流，谢谢
"If you are doing things we are interested in, and on the rails / front-end developers are confident, please send e-mail to communicate ghostm55#gitcafe.com appointment time, after determining the direct exchange of visits to our studio, thank you"
I need to communicate with an API that requires the request to be encoded in Blowfish and Base64.
In my custom library I start off with:
# encoding: utf-8
require "base64"
require 'crypt/blowfish'
I create an instance:
#blowfish_key = '1234567887654321'
#blowfish = Crypt::Blowfish.new(#blowfish_key)
And further down I create the encrypted string (or 'ticket' as the API calls it)
#string_to_encrypt = "#{#partnerid},#{user.id},#{exam_id},#{return_url},#{time_stamp}"
#enc = #blowfish.encrypt_string(#string_to_encrypt)
In the Rails console I can decrypt with #blowfish.decrypt_string(#enc) without any problems. But the API gives me gibberish:
Invalid ticket:
Decrypted String :)IŠkó}*Ogû…xÃË-ÖÐHé%q‹×ÎmªÇjEê !©†xRðá=Ͳ [À}=»ïN)'sïƒJJ=:›õ)¦$ô1X¢
Also, when I encrypt something simple in the console, like "Hello", and feed the encrypted string to an online Blowfish decoder, like http://webnet77.com/cgi-bin/helpers/blowfish.pl, I get the same gibberish mess back.
It's like the Ruby blowfish encryption is a format that is not used anywhere else.
Note:
In my actual application I send the encrypted string via a form field to the Webservice. The encrypted string is Base64 encoded and prefixed with an 'a'.
#enc = Base64.encode64(#enc)
#enc = 'a' + CGI::escape(#enc)
This is explained in their documentation:
Format of the ticket:
t=’a’ + URL_Encode (
Base64_Encode (
Blowfish_Encrypt ( ‘partnerid,user_id,test_id,URL_Encode(return_URL),ticket_timestamp’, ‘blowfish_key’)
)
)
Note above that the Blowfish_Encrypt function accepts two parameters-
1)the string to encrypt and 2)a hex key. Also note that the ticket has
been prefixed with a lower case ‘a’ (ASCII 97).
Example of what the HTML form will look like:
<form method=”POST” action=”http://www.expertrating.com/partner_site_name/”>
<input type=”hidden” name=”t” value=”adfinoidfhdfnsdfnoihoweirhqwdnd2394yuhealsnkxc234rwef45324fvsdf2” />
<input type=”submit” name=”submit” value=”Proceed to take the Test” />
</form>
I am lost, where do I go wrong?
There's a couple of things:
1) As was said in the comments, if you print the encrypted strings, then this is almost always causing trouble because the encrypted strings are very likely to contain non-ASCII characters that will either be unprintable or in a representation that is not understood by others. The best way to achieve a representation that is widely understood is to encode the result using Base64 or Hex encoding, so the Base64 encoding you apply in your application is fine for that.
2) The Perl app you linked to uses for example hex encoding. As a consequence, it will also only accept encrypted strings in hex encoding. That's why it wouldn't accept any of your inputs at all. You get hex encoding suitable for the application as follows:
hex = encrypted.unpack("H*")[0].upcase
3) But still no luck with the Perl app. One reason is this (taken from Crypt sources):
def encrypt_stream(plainStream, cryptStream)
initVector = generate_initialization_vector(block_size() / 4)
chain = encrypt_block(initVector)
cryptStream.write(chain)
What this means is that Crypt writes the IV as the first block of the encrypted message. This is totally fine, but most modern crypto libraries I know won't prepend the IV but rather assume it is exchanged as out-of-band information between the two communicating parties.
4) But even knowing this, you will still have no luck with the Perl app. The reason is that the Perl app uses ECB mode encryption where the Crypt gem uses CBC mode. CBC mode uses an IV, so even one-block messages won't match except if you are using an all-zero IV. But using an all-zero IV (or any other deterministic IV) is bad practice (and Crypt doesn't do it anyway). Doing so allows distinguishing the first block from random and opens you up to attacks like BEAST. Using ECB is bad practice as well except for totally rare edge cases. So let's forget about that Perl app and concentrate on the things at hand.
5) I'm naturally in favor of using Ruby OpenSSL, but in this case I think I'm not being subjective if I tell you it's better for overall security to use it instead of the Crypt gem. Just telling you so would be lame, so here's two reasons why:
Crypt generates its IVs using a predictable random generator (a combination of srand and rand), but that's not good enough. It has to be a cryptographically secure random generator.
Crypt seems to be no longer maintained and there have been some things going on in the meantime that either were unknown at the time or that were never in the scope of that project. For example, OpenSSL starts to deal with leakage-resilient cryptography to prevent side-channel attacks that target timing, cache misses etc. That has probably never been the intention of Crypt, but such attacks pose a real threat in real life.
6) If my preaching has convinced you to make the change, then I could continue and ask you whether it really has to be Blowfish. The algorithm itself is outstanding, no doubt, but there are better, even more secure options available now, such as AES for example. If it absolutely has to be Blowfish, it's supported by Ruby OpenSSL as well:
cipher = OpenSSL::Cipher.new('bf-cbc')
7) If your production key looks like the one in the example, then there's another weak spot. There's not enough entropy in such strings. What you should do is again use a cryptographically secure random generator to generate your key, which is quite easy in Ruby OpenSSL:
key = cipher.random_key
The nice thing is it will automatically choose an appropriate key length depending on the cipher algorithm to be used.
8) Finally, am I right in assuming that you use that encrypted result as some form of authentication token? As in you append it to the HTML being rendered, wait to receive it back in some POST request and compare the received token with the original in order to authenticate some action? If this is the case, then this would again be bad practice. This time even more so. You are not using authenticated encryption here, which means that your ciphertext is malleable. This implies that an attacker can relatively easily forge the contents of that token without actually knowing your encryption key. This leads to all sorts of attacks, even leading to total compromise involving key recovery.
You must either use authenticated encryption modes (GCM, CCM, EAX...) or use a message authentication code to detect if the ciphertexts had been tampered with. Even better, don't use encryption at all and generate your tickets by using a secure hash function. The key here is to compute the hash of a securely randomized value, otherwise it is again possible to predict the outcome. A timestamp, as used in your example, is not enough. It has to be a cryptographically secure nonce, probably generated by SecureRandom.
But then you still have to consider replay of such tokens, token hijacking, ... you see, it's not that easy.
We can't for the life of us figure this out. We need to make ColdFusion encrypt data which ruby will decrypt. We've tried so many different settings on the ColdFusion side, looked through SO posts, looked through Adobe docs, and cannot make it work. ColdFusion needs to encrypt it so ruby can do this:
aes = OpenSSL::Cipher::Cipher.new('aes-256-cbc').encrypt
aes.key = Digest::MD5.hexdigest("#{password}#{salt}")
aes.iv = Digest::MD5.hexdigest("#{salt}#{password}")[0,16]
encrypted = aes.update(data) + aes.final
ColdFusion pseudo code
key = tobase64(binaryDecode(lcase(hash(password & salt, "md5")), "hex"))
iv = lcase(left(hash(salt & password, "md5"), 16))
encrypt(data, key, "AES/CBC/PKCS5Padding", "Base64", iv)
Tried with/without the tobase64/binaryDecode (saw somebody mention that it would handle conversion back internally or something stupid). lcase is to make it generate MD5s that look like what ruby builds.
What are we doing wrong? Endless bad decrypt on the ruby side
What are we doing wrong?
You are not being careful with encodings.
You must take encodings into account.
In ColdFusion, you must only use a byte-array as key or IV, and you must only encrypt byte-arrays.
Do not deal with keys, IVs, or cleartexts in any form other than byte-array. Do not deal with them as base64-encoded strings, UTF-16 strings (what Java does by default), or any other form. You must always deal only with byte-arrays, and you must always know the encoding and use the same encoding between ColdFusion and Ruby.
You can get a byte-array from a string using an encoding. I would tend to use the UTF-8 encoding. Look at the CharsetEncode and CharsetDecode functions.
You are also using keys and IVs wrong. Keys may be generated from passwords using an algorithm such as PBKDF2, but only if you don't have a good way of generating with a cryptorandom PRNG and storing them. IVs should be generated with a cryptorandom PRNG, and may be prepended to the ciphertext when you store or transmit it as a convenient method of storing/transmitting the IV too.